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**Proof of FLT**

Where we're talking about X^N+Y^N=Z^N with N>2 of course... you don't need me to set out what FLT's is...

*First point: you can divide z^n by (x-y) with x^n-1 + yx^n-2 + y^2x^n-3 +... y^n-1

*Now you can divide it again giving z^n/(x-y)^2 and the other side ix x^n-2 + 2yx^n-3 + 3y^2x^n-4 + ... with a remainder ny^n-1/(x-y)

*Now you can multiply the RHS by a factor to produce an integer and the remainder by the same amount BUT the factor is less (usually) then (x-y).

*Therefore y and z must have co factors - which contradicts initial supposition (x,y,z coprime)

*However this fails if (x-y) = r^n (r = a combination of prime factors of z)

*However there is more!

*You can represent any solution to the pytahgorean equation by 3 numbers x^2 + y^2, x^2-y^2, and 2xy

*where x and y are aribtrary coprime numbers of opposite parity.

*You can then take the 2/n th route of this

*(x^2+y^2)^2/n, (x^2-y^2)^2/n and 2xy ^2/n so that these numbers when raised to the nth power satsify the pytahgorean equation but are also of the form a^n-b^n = c^n

*Then you can factorise (x^2 + y^2)^2/n - (x^2-y^2)^2/n which must be a factor of 4x^2y^2

*If you take out 1/x^2/n you get (1 + (y^2/x^2))^2/n) - (1-(y^2/x^2))^2/n)

*You can now do a binomial expansion and extract y^2 as a common factor and also a factor of x^(2/n) which means that the exception to the rule mentioned above cannot happen.

*hence : no solutions to x^n+y^n = x^n for n = prime>2